Reputation
1,617
Top tag
Next privilege 2,000 Rep.
Edit questions and answers
Badges
8 27
Newest
 Nice Answer
Impact
~30k people reached

Feb
1
accepted Conditional probability and almost sure equality confusion
Jan
30
awarded  Nice Answer
Jan
25
comment Conditional probability and almost sure equality confusion
Well I follow your reasoning, which means I can now prove my original statement (with the added measurability), yet I feel I am no closer to the understanding. Is there anything more you could say on this? I suppose the intuition is what I largely lack and what makes me feel like I don't actually understand any of it.
Jan
25
comment Conditional probability and almost sure equality confusion
@sinbadh I agree that if $X$ is $\mathcal G$-measurable, we get that equality. But I'm talking about $Y$ - if $Y$ is $\mathcal G$-measurable, I don't see how we get the equality $Y\stackrel {a.s.} = E^{\mathcal G} X$directly from the definition of conditional expectance.
Jan
25
comment Conditional probability and almost sure equality confusion
Ah! We might be gettting somehwere. Or at least I am, you already are "somewhere". In an hour I'll sit down with this and see if I can now clear the mess in my mind.
Jan
25
revised Conditional probability and almost sure equality confusion
added 60 characters in body
Jan
25
comment Conditional probability and almost sure equality confusion
@BCLC: Well I didn't, by "is such that" I only wanted to include the (what seemed to me as) relevant part of the definition, keeping in mind that we all know that it has to be measurable. I'll add that to the question.
Jan
25
comment Conditional probability and almost sure equality confusion
Thanks. This might be the problem. When you say $Y=Z$ in the $P$-a.s. sense in relation to $\mathcal G$, does this mean that in the context a.s. means with relation to $\mathcal G$ and not $\mathcal F$? I don't see why.
Jan
25
comment Conditional probability and almost sure equality confusion
Thanks, that clears some of the confusion (and possibly answers this question). Could you elaborate on why the measurability is sufficient?
Jan
25
revised Conditional probability and almost sure equality confusion
added 6 characters in body
Jan
25
asked Conditional probability and almost sure equality confusion
Jan
10
accepted Limit of a random walk with zero mean
Jan
10
comment Limit of a random walk with zero mean
Ah, dammit. Well, thank you :)
Jan
10
asked Limit of a random walk with zero mean
Dec
25
comment Is it really true that the Cartesian product $\mathbb R^2 \times \mathbb R^3$ is not equal to $\mathbb R^5$?
@ZevChonoles True, I will think about that from now on. Any suggestions for a better title?
Dec
25
comment Is it really true that the Cartesian product $\mathbb R^2 \times \mathbb R^3$ is not equal to $\mathbb R^5$?
@AsafKaragila Ah, I didn't think of seeing it as an problem with the associative property! Thanks
Dec
25
comment Is it really true that the Cartesian product $\mathbb R^2 \times \mathbb R^3$ is not equal to $\mathbb R^5$?
Which is another reason why this will be an embarrassement. The problem is, how do you search for things like this? I tried a couple of searches but couldn't find anything. Perhaps improving my searching skills rather than set theoretic knowledge is what I am after :)
Dec
25
asked Is it really true that the Cartesian product $\mathbb R^2 \times \mathbb R^3$ is not equal to $\mathbb R^5$?
Dec
24
comment Past open problems with sudden and easy-to-understand solutions
This is great! Showing this in an integral calculus course would improve the students' self confidence for sure.
Dec
23
comment Importance of Locally Compact Hausdorff Spaces
Haha, you're right about my guess, if math doesn't work out, I'll go into politics. Thank you, will read the answer(s) properly once this christmas madness passes.